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Let [tex]\( z_1 = 1 + i \)[/tex] and [tex]\( z_2 = \sqrt{3} - i \)[/tex].

Find the four fourth roots of [tex]\(-16\)[/tex].


Sagot :

To find the four fourth roots of [tex]\( -16 \)[/tex], we will follow these steps:

### Step 1: Express [tex]\(-16\)[/tex] in Polar Form
A complex number can be expressed in polar form as:
[tex]\[ z = re^{i\theta} \][/tex]
where [tex]\( r \)[/tex] is the modulus (magnitude) and [tex]\( \theta \)[/tex] is the argument (angle).

For the number [tex]\(-16\)[/tex], we identify the following:
- The modulus [tex]\( r \)[/tex] is the distance from the origin to the point, which is [tex]\( 16 \)[/tex] (since [tex]\( |-16| = 16 \)[/tex]).
- The argument [tex]\( \theta \)[/tex] is the angle in radians. Since [tex]\(-16\)[/tex] lies on the negative real axis, directly left of the origin, its argument is [tex]\( \pi \)[/tex] radians (or 180 degrees).

So we can write:
[tex]\[ -16 = 16e^{i\pi} \][/tex]

### Step 2: Find the Fourth Roots
The fourth roots of a complex number can be found using the formula:
[tex]\[ z_k = r^{1/4} e^{i(\theta + 2k\pi)/n} \][/tex]
where [tex]\( r \)[/tex] is the modulus, [tex]\( \theta \)[/tex] is the argument, [tex]\( n \)[/tex] is the degree of the root (4th root here), and [tex]\( k = 0, 1, 2, 3 \)[/tex] are the different roots.

Given:
[tex]\[ r = 16 \][/tex]
[tex]\[ \theta = \pi \][/tex]
[tex]\[ n = 4 \][/tex]

### Step 3: Calculate Each Root

For [tex]\( k = 0 \)[/tex]:
[tex]\[ z_0 = 16^{1/4} e^{i(\pi + 2 \cdot 0 \cdot \pi)/4} = 16^{1/4} e^{i\pi/4} \][/tex]
[tex]\[ 16^{1/4} = \sqrt{2} \cdot \sqrt{2} = \sqrt{4} = 2 \][/tex]
[tex]\[ z_0 = 2 e^{i\pi/4} = 2 \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 2 \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = \sqrt{2} + i\sqrt{2} \][/tex]

For [tex]\( k = 1 \)[/tex]:
[tex]\[ z_1 = 16^{1/4} e^{i(\pi + 2 \cdot 1 \cdot \pi)/4} = 16^{1/4} e^{i3\pi/4} \][/tex]
[tex]\[ z_1 = 2 e^{i3\pi/4} = 2 \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) = 2 \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = -\sqrt{2} + i\sqrt{2} \][/tex]

For [tex]\( k = 2 \)[/tex]:
[tex]\[ z_2 = 16^{1/4} e^{i(\pi + 2 \cdot 2 \cdot \pi)/4} = 16^{1/4} e^{i5\pi/4} \][/tex]
[tex]\[ z_2 = 2 e^{i5\pi/4} = 2 \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = 2 \left( -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = -\sqrt{2} - i\sqrt{2} \][/tex]

For [tex]\( k = 3 \)[/tex]:
[tex]\[ z_3 = 16^{1/4} e^{i(\pi + 2 \cdot 3 \cdot \pi)/4} = 16^{1/4} e^{i7\pi/4} \][/tex]
[tex]\[ z_3 = 2 e^{i7\pi/4} = 2 \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right) = 2 \left( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = \sqrt{2} - i\sqrt{2} \][/tex]

### Summary of Roots
So, the four fourth roots of [tex]\(-16\)[/tex] are:
[tex]\[ \sqrt{2} + i\sqrt{2}, \][/tex]
[tex]\[ -\sqrt{2} + i\sqrt{2}, \][/tex]
[tex]\[ -\sqrt{2} - i\sqrt{2}, \][/tex]
[tex]\[ \sqrt{2} - i\sqrt{2}. \][/tex]